$\displaystyle (a - b)^3 - b^3$

$\displaystyle [(a - b) - b][(a - b)^2 + (a - b)b + b^2]$

Is that right?

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- Feb 29th 2008, 05:06 PMMadSpecial Factoring
$\displaystyle (a - b)^3 - b^3$

$\displaystyle [(a - b) - b][(a - b)^2 + (a - b)b + b^2]$

Is that right? - Feb 29th 2008, 05:15 PMSoroban
Correct, Mad!

. . . Nice going!

- Feb 29th 2008, 05:50 PMMad
Gah, this one is driving me nuts :

Factor:

$\displaystyle 27a^4 - a$

$\displaystyle a(27a^3 - 1)$

Or can I do more? - Feb 29th 2008, 05:55 PMmr fantastic
- Feb 29th 2008, 05:56 PMMad
I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks. (Headbang)

EDIT: Just in case...

$\displaystyle (a)(27a - 1)(27a^2 + 27a + 1)$ - Feb 29th 2008, 06:16 PMmr fantastic
- Mar 3rd 2008, 06:17 PMMad
Sorry, I don't understand. I looked back over my notes but I'm kind of confused. (Thinking)

- Mar 3rd 2008, 06:36 PMtopsquark
$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3$

and

$\displaystyle x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

So let $\displaystyle x = 3a$ and $\displaystyle y = 1$

Thus

$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)$

$\displaystyle = (3a - 1)(9a^2 + 3a + 1)$

-Dan - Mar 3rd 2008, 06:37 PMMad
Oh, now I understand, thank you. (Smile)